Chapter 1, Problem 11P

What, roughly, is the percent uncertainty in the volume of a spherical beach ball of radius $r=0.84\pm 0.04$ m?

To determine

The roughpercent uncertainty in the volume of a spherical beach ball of radius, $r=0.84\pm 0.04\text{ m}.$

Answer to Problem 11P

Solution:

  $14.1\text{\%}$

Explanation of Solution

Given:

Spherical beach ball of radius $r=0.84\pm 0.04\text{ m}.$

  $\begin{array}{l}\text{Absolute radius = 0}\text{.84 m}\hfill \\ \text{Radius uncertainty = ±0}\text{.04 m}\hfill \end{array}$

Formula Used:

First, Absolute Volume, V is found using the formula

  $V=\left(\frac{4}{3}\right)\times \pi \times r^3$ .

Where, r is the radius.

Uncertainty in volume is found using the formula $\left(\frac{\Delta V}{V}\right)=\pm n\left(\frac{\Delta r}{r}\right)$

Where, n is the power to the radius in the formula.

Example − If in formula $r^2$ then $n=2$

Calculations:

  $\begin{array}{l}V=\left(\frac{4}{3}\right)\times \pi \times r^3\\ =\left(\frac{4}{3}\right)\times 3.14\times {\left(0.84\right)}^3\\ =2.48{\text{m}}^{\text{3}}\end{array}$

  $\begin{array}{l}\left(\frac{\Delta V}{V}\right)=\pm 3\left(\frac{\Delta r}{r}\right)\\ \Rightarrow \left(\frac{\Delta V}{V}\right)=\pm 3\left(\frac{0.04}{0.84}\right)\\ \Rightarrow \left(\frac{\Delta V}{V}\right)=\pm 0.14\\ \Delta V=\pm 0.35\end{array}$

  $\text{Volume}=2.48\pm 0.35{\text{m}}^{\text{3}}$

Hence, percent uncertainty in volume is found using

  $\frac{\left(\text{Volume Uncertainty}\right)}{\left(\text{Absolute Volume}\right)}\text{ ×100\% }$

  $\begin{array}{l}\text{Here, Absolute Volume is 2}{\text{.48 m}}^{\text{3}}\text{.}\hfill \\ \text{Volume Uncertainty is ±0}{\text{.35 m}}^{\text{3}}\text{.}\hfill \end{array}$

  $\begin{array}{l}\text{Percent Uncertainty}=\frac{0.35}{2.48}\times 100\%\\ \text{Percent Uncertainty}=0.141\times 100\%\end{array}$

  $\text{Percent Uncertainty}=14.1\%$

Conclusion:

Percent uncertainty in the volume of a spherical beach ball is $14.1\text{\%}$ .